What is the equation of the normal line of #f(x)= x^2/(x^3− 5x + 1)# at #x = 5#?

Question

Isabella Ackerman · Accepted Answer

Equation of normal line is # y= 13.79x - 68.7# #f(x)=x^2/(x^3-5x+1) :. f(5)=5^2/(5^3-5*5+1) ~~0.25(2dp) # Point is at #(x_1,y_1) =(5,0.25)# . Slope is #m= f'(x)# #f'(x) = (2x*(x^3-5x+1)- x^2(3x^2-5))/(x^3-5x+1)^2# #f'(x) =m= (25*(5^3-5*5+1)- 5^2(3*5^2-5))/(5^3-5*5+1)^2# or #m ~~-0.07 #. Slope of normal is #m_1= -1/-0.07~~13.79(2dp)# Equation of normal line is #(y-y_1)=m_1(x-x_1)# or #(y-0.25)=13.79(x-5) or y= 13.79x - 68.7# [Ans]

Cameron Alexander · Answer

undefined To find the equation of the normal line of f(x) = x^2/(x^3 - 5x + 1) at x = 5, we need to determine the slope of the tangent line at x = 5 and then find the negative reciprocal of that slope to obtain the slope of the normal line.

To find the slope of the tangent line at x = 5, we differentiate f(x) with respect to x and evaluate it at x = 5.

Differentiating f(x) = x^2/(x^3 - 5x + 1) using the quotient rule, we get:

f'(x) = (2x(x^3 - 5x + 1) - x^2(3x^2 - 5))/(x^3 - 5x + 1)^2

Simplifying this expression, we have:

f'(x) = (2x^4 - 10x^2 + 2x - 3x^4 + 5x^2)/(x^3 - 5x + 1)^2

Combining like terms, we get:

f'(x) = (-x^4 - 5x^2 + 2x)/(x^3 - 5x + 1)^2

Now, we can evaluate f'(x) at x = 5:

f'(5) = (-5^4 - 5(5)^2 + 2(5))/(5^3 - 5(5) + 1)^2

Simplifying this expression, we have:

f'(5) = (-625 - 125 + 10)/(125 - 25 + 1)^2

f'(5) = (-740)/(101)^2

f'(5) = -0.072871

The slope of the tangent line at x = 5 is -0.072871.

To find the slope of the normal line, we take the negative reciprocal of the slope of the tangent line:

Slope of normal line = -1/(-0.072871) = 13.725

Now, we have the slope of the normal line. To find the equation of the normal line, we use the point-slope form of a line and substitute the values of x = 5 and the slope:

y - y1 = m(x - x1)

Using the point (5, f(5)) on the curve, we substitute x = 5 and y = f(5) into the equation:

y - f(5) = 13.725(x - 5)

Simplifying this equation, we have:

y - f(5) = 13.725x - 68.625

y = 13.725x - 68.625 + f(5)

Therefore, the equation of the normal line of f(x) = x^2/(x^3 - 5x + 1) at x = 5 is y = 13.725x - 68.625 + f(5).